Combustion of 1 mole of benzene is expressed at C6H6(1) + 15/2 O2(g) → CO2(g) + 3H2O(1).

Question

Combustion of 1 mole of benzene is expressed at

\(C_6H_6(1) + \frac {15}{2}O_2 (g) \rightarrow CO_2 (g) + 3H_2 O (1)\)

The standard enthalpy of combustion of 2 mol of benzene is – 'x' kJ. x = _____.

(1) standard Enthalpy of formation of 1 mol of \(C_6 H_6 (1)\), for the reaction

6C(graphite) + \(3H_2(g) \rightarrow C_6H_6(1)\) is 48.5 kJ mol^-1.

(2) Standard Enthalpy of formation of 1 mol of \(CO_2 (g)\), for the reaction

C(graphite) + \(O _{2(g)} \rightarrow CO_2 (g)\) is -393.5 kJ \(mol^{-1 }\).

(3) Standard and Enthalpy of formation of 1 mol of \(H_2O(1)\), for the reaction

\(H_2(g) + \frac {1}{2}O_2 (g) \rightarrow H_2O (1)\) is \(-286 \, kJ\,mol ^{-1}.\)

Answer 1

Correct answer is : 6535

\( 6 \mathrm{C} \text { (graphite) }+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{C}_6 \mathrm{H}_6(\ell) ; \Delta \mathrm{H}=48.5 \mathrm{~kJ} / \mathrm{mol} \)

\(\mathrm{C} \text { (graphite) }+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta \mathrm{H}=-393.5 \mathrm{~kJ} / \mathrm{mol} \)

\(\mathrm{H}_2^{(\mathrm{g})}+\frac{1}{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\ell) ; \Delta \mathrm{H}=-286 \mathrm{~kJ} / \mathrm{mol}\)

equation -(1) x 1 + (2) x 6 + (3) x 3 - 48.5 -6 x 393.5 - 3 x 286 

= - 3267.5 kJ for 1 mol 

= - 6535 kJ for 2 mol